Chapter 5, Problem 5P

In mice, the dominant allele Gs of the X-linked gene Greasy produces shiny fur, while the recessive wild-type Gs⁺ allele determines normal fur. The dominant allele Bhd of the X-linked Broadhead gene causes skeletal abnormalities including broad heads and snouts, while the recessive wild-type Bhd⁺ allele yields normal skeletons. Female mice heterozygous for the two alleles of both genes were mated with wild-type males. Among 100 male progeny of this cross, 49 had shiny fur, 48 had skeletal abnormalities, 2 had shiny fur and skeletal abnormalities, and 1 was wild type.

a.	Diagram the cross described and calculate the distance between the two genes.
b.	What would have been the results if you had counted 100 female progeny of the cross?

Summary Introduction

a.

To draw:

The diagram of the cross described in the problem and calculate the distance between the two genes.

Introduction:

Genes present on the same chromosome that do not assort independently are said to be linked. Recombination is a process in which small DNA sequences are broken and then recombined to give new combinations of alleles. The genetic diversity at the level of genes is created by the recombination. This reflects variation in the DNA sequences of different organisms.

Explanation of Solution

In mice, the dominant allele Gs of the X-linked gene Greasy gives shiny fur. The recessive wild-type Gs⁺ allele gives normal fur. The dominant allele Bhd of the X-linked Broadhead gene causes skeletal abnormalities. The recessive wild-type Bhd⁺ allele gives a normal skeleton. The heterozygous female mice for the two alleles of both genes were mated with wild-type males.

The genotype of heterozygous female mice for the two alleles of both genes: $X^{Gs/Bh{d}^{+}}{X}^{G{s}^{+}/Bhd}$

The genotype of wild type male: $X^{G{s}^{+}/Bh{d}^{+}}Y$

The following table explains the cross between the heterozygous female mice for the two alleles of both genes and the wild type male:

Male/Female gametes	$X^{G{s}^{+}/Bh{d}^{+}}$	$X^{G{s}^{+}/Bhd}$	$X^{Gs/Bhd}$	$X^{Gs/Bh{d}^{+}}$
Y	$X^{G{s}^{+}/Bh{d}^{+}}Y$ (1 wild type)	$X^{G{s}^{+}/Bhd}Y$ (49 shiny fur)	$X^{Gs/Bhd}Y$ (2 shiny fur, skeletal abnormalities)	$X^{Gs/Bh{d}^{+}}Y$ (48 skeletal abnormalities)

Here, 49 and 48 male progenies are parental gametes; whereas, 1 and 2 are recombinant gametes. Recombination frequency (RF) is the rate of occurrence of recombination between a pair of linked genes. It can be calculated using the following formula:

$RF=\left(\frac{\text{Total number of recombinant}}{\text{Total number of progeny}}\right)$

Assuming, 1 map unit is equal to 1% recombination. Recombination frequency helps in determining the distance between two genes and in generating linkage map.

So,

$\begin{array}{rll}RF& =& \left(\frac{1+2}{49+1+2+48}\right)\times 100\\ & =& \left(\frac{3}{100}\right)\times 100\\ & =& 3\%\text{or m}\text{.u or cM}\end{array}$

Therefore, the distance between the given genes is 3 cM.

Summary Introduction

b.

To determine:

The results if 100 female progeny would have counted.

Introduction:

Recombination frequency (RF) is the rate of occurrence of recombination between a pair of linked genes. It helps in determining the distance between two genes and in the generation of the linkage map.

Explanation of Solution

The genotype of heterozygous female mice for the two alleles of both genes: $X^{Gs/Bh{d}^{+}}{X}^{G{s}^{+}/Bhd}$

The genotype of wild type male: $X^{G{s}^{+}/Bh{d}^{+}}Y$

The following table explains the cross between the heterozygous female mice for the two alleles of both genes and the wild type male:

Male/Female gametes	$X^{G{s}^{+}/Bh{d}^{+}}$	$X^{G{s}^{+}/Bhd}$	$X^{Gs/Bhd}$	$X^{Gs/Bh{d}^{+}}$
$X^{G{s}^{+}/Bh{d}^{+}}$	$X^{G{s}^{+}/Bh{d}^{+}}{X}^{G{s}^{+}/Bh{d}^{+}}$ (1 wild type)	$X^{G{s}^{+}/Bh{d}^{+}}{X}^{G{s}^{+}/Bhd}$ (48 skeletal abnormalities)	$X^{G{s}^{+}/Bh{d}^{+}}{X}^{Gs/Bhd}$ (2 shiny fur, skeletal abnormalities)	$X^{G{s}^{+}/Bh{d}^{+}}{X}^{Gs/Bh{d}^{+}}$ (49 shiny fur)

So, in this case, the same number of progeny would be obtained, even 100 female progeny would have been counted.

Therefore, the distance between the two given genes would be 3 cM.